PPT Probability ditribution basics presentation

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Statistics for Business and Economics (13e)
Welcome to Probability: Basic, Discrete & Continuous

2
Probability:
Probability is a numerical
measure of the likelihood
that an event will occur.
Probability values are
always assigned on a scale
from 0 to 1.
Why do we need
Probability?
Decision making in
uncertain business
environment.
It’s Key to any
Business
Life is uncertain and full of surprise. Do you
know what will happen tomorrow?
Make decision and live with the
consequence.
---Anonymous Probability Teacher 

3
Statistics for
Business and Economics (13e)
Anderson, Sweeney, Williams, Camm, Cochran
© 2017 Cengage Learning
Slides by John Loucks
St. Edwards University

4
Chapter 4
Introduction to Probability
• Experiments, Counting Rules, and Assigning Probabilities
• Events and Their Probability
• Some Basic Relationships of Probability
• Conditional Probability
• Bayes’ Theorem

5
Uncertainties
• Managers often base their decisions on an analysis of uncertainties such
as the following:
• What are the chances that sales will decrease if we increase prices?
• What is the likelihood a new assembly method will increase
productivity?
• What are the odds that a new investment will be profitable?

6
Probability
• Probability is a numerical measure of the likelihood that an event will
occur.
• Probability values are always assigned on a scale from 0 to 1.
• A probability near zero indicates an event is quite unlikely to occur.
• A probability near one indicates an event is almost certain to occur.

7
Probability as a Numerical Measure
of the Likelihood of Occurrence
0 1
.5
Increasing Likelihood of Occurrence
Probability:
The event
is very
unlikely
to occur.
The occurrence
of the event is
just as likely as
it is unlikely.
The event
is almost
certain
to occur.

8
Statistical Experiments
• In statistics, the notion of an experiment differs somewhat from that of
an experiment in the physical sciences.
• In statistical experiments, probability determines outcomes.
• Even though the experiment is repeated in exactly the same way, an
entirely different outcome may occur.
• For this reason, statistical experiments are sometimes called random
experiments.

9
An Experiment and Its Sample Space
• An experiment is any process that generates well-defined outcomes.
• The sample space for an experiment is the set of all experimental outcomes.
• An experimental outcome is also called a sample point.

10
Experiment
Toss a coin
Inspection a part
Conduct a sales call
Roll a die
Play a football game
Experiment Outcomes
Head, tail
Defective, non-defective
Purchase, no purchase
1, 2, 3, 4, 5, 6
Win, lose, tie

11
Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley
has determined that the possible outcomes of these investments three
months from now are as follows.
Investment Gain or Loss
in 3 Months (in $1000s)
Markley Oil Collins Mining
10
5
0
-20
8
-2
• Example: Bradley Investments

12
A Counting Rule for Multiple-Step Experiments
• If an experiment consists of a sequence of k steps in which there are n1
possible results for the first step, n2 possible results for the second step, and
so on, then the total number of experimental outcomes is given by (n1)
(n2) . . . (nk).
• A helpful graphical representation of a multiple-step experiment is a tree
diagram.

13
• Bradley Investments can be viewed as a two-step experiment. It involves
two stocks, each with a set of experimental outcomes.
Markley Oil: n1 = 4
Collins Mining: n2 = 2
Total Number of
Experimental Outcomes: n1n2 = (4)(2) = 8
A Counting Rule for Multiple-Step Experiments

14
Tree Diagram
Gain 5
Gain 8
Gain 8
Gain 10
Gain 8
Gain 8
Lose 20
Lose 2
Lose 2
Lose 2
Lose 2
Even
Markley Oil
(Stage 1)
Collins Mining
(Stage 2)
Experimental
Outcomes
(10, 8) Gain $18,000
(10, -2) Gain $8,000
(5, 8) Gain $13,000
(5, -2) Gain $3,000
(0, 8) Gain $8,000
(0, -2) Lose $2,000
(-20, 8) Lose $12,000
(-20, -2) Lose $22,000

15
Counting Rule for Combinations
• A second useful counting rule enables us to count the number of
experimental outcomes when n objects are to be selected from a set of
N objects.
• Number of Combinations of N Objects Taken n at a Time
where: N! = N(N - 1)(N - 2) . . . (2)(1)
n! = n(n - 1)(n - 2) . . . (2)(1)
0! = 1
=

16
• Number of Permutations of N Objects Taken n at a Time
where: N! = N(N - 1)(N - 2) . . . (2)(1)
n! = n(n - 1)(n - 2) . . . (2)(1)
0! = 1
Counting Rule for Permutations
• A third useful counting rule enables us to count the number of
experimental outcomes when n objects are to be selected from a set of
N objects, where the order of selection is important.
=

17
Assigning Probabilities
• Basic Requirements for Assigning Probabilities
1. The probability assigned to each experimental outcome must be
between 0 and 1, inclusively.
0 < P(Ei) < 1 for all i
where: Ei is the i th experimental outcome
and P(Ei) is its probability

18
2. The sum of the probabilities for all experimental outcomes must equal 1.
P(E1) + P(E2) + . . . + P(En) = 1
where: n is the number of experimental outcomes
• Basic Requirements for Assigning Probabilities

19
• Classical Method
• Relative Frequency Method
• Subjective Method
Assigning probabilities based on the assumption of equally likely
outcomes
Assigning probabilities based on experimentation or historical data
Assigning probabilities based on judgment

20
Classical Method
If an experiment has n possible outcomes, the classical method would
assign a probability of 1/n to each outcome.
Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a 1/6 chance of occurring
• Example: Rolling a Die

21
Relative Frequency Method
Number of
Polishers Rented
Number
of Days
0
1
2
3
4
4
6
18
10
2
Lucas Tool Rental would like to assign probabilities to the number of car
polishers it rents each day. Office records show the following frequencies of
daily rentals for the last 40 days.
• Example: Lucas Tool Rental

22
Each probability assignment is given by dividing the frequency (number
of days) by the total frequency (total number of days).
Probability
Number of
Polishers Rented
Number
of Days
0
1
2
3
4
4
6
18
10
2
40
.10 = 4/40
.15
.45
.25
.05
1.00
Relative Frequency Method
• Example: Lucas Tool Rental

23
Subjective Method
• When economic conditions or a company’s circumstances change rapidly it
might be inappropriate to assign probabilities based solely on historical data.
• We can use any data available as well as our experience and intuition, but
ultimately a probability value should express our degree of belief that the
experimental outcome will occur.
• The best probability estimates often are obtained by combining the estimates
from the classical or relative frequency approach with the subjective estimate.

24
Subjective Method
An analyst made the following probability estimates.
Exper. Outcome Net Gain or Loss Probability
(10, 8)
(10, -2)
(5, 8)
(5, -2)
(0, 8)
(0, -2)
(-20, 8)
(-20, -2)
$18,000 Gain
$8,000 Gain
$13,000 Gain
$3,000 Gain
$8,000 Gain
$2,000 Loss
$12,000 Loss
$22,000 Loss
.20
.08
.16
.26
.10
.12
.02
.06
1.00

25
• An event is a collection of sample points.
• The probability of any event is equal to the sum of the probabilities of the
sample points in the event.
• If we can identify all the sample points of an experiment and assign a
probability to each, we can compute the probability of an event.
Events and Their Probabilities

26
Event M = Markley Oil Profitable
M = {(10, 8), (10, -2), (5, 8), (5, -2)}
P(M) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2)
= .20 + .08 + .16 + .26
= .70

27
Event C = Collins Mining Profitable
C = {(10, 8), (5, 8), (0, 8), (-20, 8)}
P(C) = P(10, 8) + P(5, 8) + P(0, 8) + P(-20, 8)
= .20 + .16 + .10 + .02
= .48

28
Some Basic Relationships of Probability
• There are some basic probability relationships that can be used to compute the
probability of an event without knowledge of all the sample point probabilities.
Complement of an Event
Intersection of Two Events
Mutually Exclusive Events
Union of Two Events

29
• The complement of A is denoted by Ac
.
• The complement of event A is defined to be the event consisting of all
sample points that are not in A.
Complement of an Event
Event A Ac
Sample
Space S
Venn Diagram

30
• The union of events A and B is denoted by A  B.
• The union of events A and B is the event containing all sample points that
are in A or B or both.
Union of Two Events
Sample
Space S
Event A Event B
Venn Diagram

31
M  C = Markley Oil Profitable
or Collins Mining Profitable (or both)
M  C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)}
P(M  C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) + P(0, 8) + P(-20, 8)
= .20 + .08 + .16 + .26 + .10 + .02
= .82
Union of Two Events

32
• The intersection of events A and B is denoted by A  B.
• The intersection of events A and B is the set of all sample points that are in
both A and B.
Sample
Space S
Event A Event B
Intersection of A and B
Venn Diagram

33
M  C = Markley Oil Profitable and Collins Mining Profitable
M  C = {(10, 8), (5, 8)}
P(M  C) = P(10, 8) + P(5, 8)
= .20 + .16
= .36

34
• The addition law provides a way to compute the probability of event A, or
B, or both A and B occurring.
Addition Law
• The law is written as:
P(A  B) = P(A) + P(B) - P(A  B)

35
M  C = Markley Oil Profitable or Collins Mining Profitable
We know: P(M) = .70, P(C) = .48, P(M  C) = .36
Thus: P(M  C) = P(M) + P(C) - P(M  C)
= .70 + .48 - .36
= .82
(This result is the same as that obtained earlier
using the definition of the probability of an event.)
Addition Law

36
• Two events are said to be mutually exclusive if the events have no sample
points in common.
• Two events are mutually exclusive if, when one event occurs, the other
cannot occur.
Sample
Space S
Event A Event B
Venn Diagram

37
• If events A and B are mutually exclusive, P(A  B) = 0.
• The addition law for mutually exclusive events is:
P(A  B) = P(A) + P(B)

38
• The probability of an event given that another event has occurred is called a
conditional probability.
• A conditional probability is computed as follows :
• The conditional probability of A given B is denoted by P(A|B).
Conditional Probability
𝑃 ( 𝐴|𝐵)=
𝑃( 𝐴∩ 𝐵)
𝑃(𝐵)

39
We know: P(M  C) = .36, P(M) = .70
Thus:
P(C|M) = Collins Mining Profitable given Markley Oil Profitable
=
Conditional Probability

40
Multiplication Law
• The multiplication law provides a way to compute the probability of the
intersection of two events.
P(A  B) = P(B)P(A|B)

41
We know: P(M) = .70, P(C|M) = .5143
M  C = Markley Oil Profitable and Collins Mining Profitable
Thus: P(M  C) = P(M)P(M|C)
= (.70)(.5143)
= .36
(This result is the same as that obtained earlier
using the definition of the probability of an event.)
Multiplication Law

42
Joint Probability Table
Collins Mining
Profitable (C) Not Profitable (Cc
)
Markley Oil
Profitable (M)
Not Profitable (Mc
)
Total .48 .52
Total
.70
.30
1.00
.36 .34
.12 .18
• Joint probabilities appear in the center of the table.
• Marginal probabilities appear in the margins of the table.

43
Independent Events
• If the probability of event A is not changed by the existence of event B,
we would say that events A and B are independent.
• Two events A and B are independent if:
P(A|B) = P(A) or P(B|A) = P(B)

44
• The multiplication law also can be used as a test to see if two events are
independent.
P(A  B) = P(A)P(B)
Multiplication Law for Independent Events

45
We know: P(M  C) = .36, P(M) = .70, P(C) = .48
But: P(M)P(C) = (.70)(.48) = .34, not .36
Are events M and C independent?
Does P(M  C) = P(M)P(C) ?
Hence: M and C are not independent.
Multiplication Law for Independent Events

46
• Do not confuse the notion of mutually exclusive events with that of
independent events.
• Two events with nonzero probabilities cannot be both mutually exclusive
and independent.
• If one mutually exclusive event is known to occur, the other cannot
occur.; thus, the probability of the other event occurring is reduced to
zero (and they are therefore dependent).
Mutual Exclusiveness and Independence
• Two events that are not mutually exclusive, might or might not be
independent.

47
Bayes’ Theorem
New
Information
Application
of Bayes’
Theorem
Posterior
Probabilities
Prior
Probabilities
• Often we begin probability analysis with initial or prior probabilities.
• Then, from a sample, special report, or a product test we obtain some
additional information.
• Given this information, we calculate revised or posterior probabilities.
• Bayes’ theorem provides the means for revising the prior probabilities.

48
A proposed shopping center will provide strong competition for downtown
businesses like L. S. Clothiers. If the shopping center is built, the owner of L. S.
Clothiers feels it would be best to relocate to the shopping center.
Bayes’ Theorem
• Example: L. S. Clothiers
The shopping center cannot be built unless a zoning change is
approved by the town council. The planning board must first make a
recommendation, for or against the zoning change, to the council.

49
Let:
Prior Probabilities
A1 = town council approves the zoning change
A2 = town council disapproves the change
P(A1) = .7, P(A2) = .3
Using subjective judgment:

50
New Information
The planning board has recommended against the zoning change. Let B
denote the event of a negative recommendation by the planning board.
Given that B has occurred, should L. S. Clothiers revise the probabilities
that the town council will approve or disapprove the zoning change?

51
Past history with the planning board and the town council indicates
the following:
Conditional Probabilities
P(B|A1) = .2 and P(B|A2) = .9
P(BC
|A1) = .8 and P(BC
|A2) = .1
Hence:

52
P(Bc
|A1) = .8
P(A1) = .7
P(A2) = .3
P(B|A2) = .9
P(Bc
|A2) = .1
P(B|A1) = .2
P(A1  B) = .14
P(A2  B) = .27
P(A2  Bc
) = .03
P(A1  Bc
) = .56
Town Council Planning Board Experimental Outcomes
Tree Diagram
1.00

53
Bayes’ Theorem
• To find the posterior probability that event Ai will occur given that event B
has occurred, we apply Bayes’ theorem.
• Bayes’ theorem is applicable when the events for which we want to
compute posterior probabilities are mutually exclusive and their union is
the entire sample space.
𝑃 ( 𝐴𝑖|𝐵)=
𝑃 ( 𝐴𝑖) 𝑃 (𝐵∨ 𝐴𝑖)
𝑃 ( 𝐴1) 𝑃 (𝐵|𝐴1)+𝑃 ( 𝐴2) 𝑃 (𝐵|𝐴2)+…+𝑃 ( 𝐴𝑛) 𝑃(𝐵∨ 𝐴𝑛)

54
Posterior Probabilities
Given the planning board’s recommendation not to approve the zoning
change, we revise the prior probabilities as follows:
= .34
𝑃 ( 𝐴1
|𝐵)=
𝑃 ( 𝐴1) 𝑃 (𝐵∨ 𝐴1)
𝑃 ( 𝐴1) 𝑃 ( 𝐵|𝐴1)+𝑃 ( 𝐴2) 𝑃 (𝐵|𝐴2)
=

55
The planning board’s recommendation is good news for L. S. Clothiers.
The posterior probability of the town council approving the zoning
change is .34 compared to a prior probability of .70.
Posterior Probabilities

56
Bayes’ Theorem: Tabular Approach
Column 1 - The mutually exclusive events for which posterior probabilities
are desired.
Column 2 - The prior probabilities for the events.
Column 3 - The conditional probabilities of the new information given
each event.
Prepare the following three columns:
• Step 1

57
(1) (2) (3) (4) (5)
Events
Ai
Prior
Probabilities
P(Ai)
Conditional
Probabilities
P(B|Ai)
A1
A2
.7
.3
1.0
.2
.9
• Step 1

58
Column 4
Compute the joint probabilities for each event and the new
information B by using the multiplication law.
Prepare the fourth column:
Multiply the prior probabilities in column 2 by the corresponding
conditional probabilities in column 3. That is, P(Ai IB) = P(Ai) P(B|Ai).
• Step 2

59
• Step 2
(1) (2) (3) (4) (5)
Events
Ai
Prior
Probabilities
P(Ai)
Conditional
Probabilities
P(B|Ai)
A1
A2
.7
.3
1.0
.2
.9
.14 = .7(.2)
.27
Joint
Probabilities
P(Ai I B)

60
• Step 2 (continued)
We see that there is a .14 probability of the town council approving
the zoning change and a negative recommendation by the planning
board.
There is a .27 probability of the town council disapproving the zoning
change and a negative recommendation by the planning board.

61
Sum the joint probabilities in Column 4. The sum is the probability
of the new information, P(B). The sum .14 + .27 shows an overall
probability of .41 of a negative recommendation by the planning
board.
• Step 3

62
• Step 3
(1) (2) (3) (4) (5)
Events
Ai
Prior
Probabilities
P(Ai)
Conditional
Probabilities
P(B|Ai)
A1
A2
.7
.3
1.0
.2
.9
.14
.27
Joint
Probabilities
P(Ai I B)
P(B) = .41

63
Prepare the fifth column:
Column 5
Compute the posterior probabilities using the basic relationship of
conditional probability.
The joint probabilities P(Ai I B) are in column 4 and the probability
P(B) is the sum of column 4.
𝑃 ( 𝐴𝑖|𝐵)=
𝑃( 𝐴𝑖∩ 𝐵)
𝑃( 𝐵)
• Step 4

64
• Step 4
(1) (2) (3) (4) (5)
Events
Ai
Prior
Probabilities
P(Ai)
Conditional
Probabilities
P(B|Ai)
A1
A2
.7
.3
1.0
.2
.9
.14
.27
Joint
Probabilities
P(Ai I B)
P(B) = .41
Posterior
Probabilities
P(Ai |B)
.3415 = .14/.41
.6585
1.0000

65
End of Chapter 4

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